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xxTIMURxx [149]
3 years ago
8

Alex buys a top of the line computer. He did not realize that the computer would lose value so fast. If his computer cost $1800.

00 and it depreciates at a rate of 45% each year, in how many years will it be worth less than 1/3 of what he paid for it?
Mathematics
1 answer:
Natali [406]3 years ago
3 0

This can be solve by using the formula

D = P( 1 – i)^n

Where d is the depreciation value after n years

P is the initial value

i is the depreciation rate

n is the years

D = 1/3 ( 1800)

D = 600

So

600 = 1800 ( 1- 0.45)^n

Solve for n

<span>N = 1.83 years</span>

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Water freezes at 0 degrees Celsius. A student discovered that adding half a cup of salt to a gallon of water lowers its freezing
Artemon [7]

Answer:

-7°C

Step-by-step explanation:

Freezing point of water = 0°C

By adding salt, freezing point lowers by 7°C ;

The freezing point of salt water can be expressed as ;

Since temperature lowers :

Freezing point of salt water :

Freezing point of water - 7°C

0°C - 7°C

= - 7°C

4 0
3 years ago
A segment has an endpoint at (2, 1). The midpoint is at (5, 1). What are the coordinates of the other endpoint?
Free_Kalibri [48]
(8,1) since the midpoint is 3 away from the endpoint, it must also be 3 away from the other endpoint
7 0
3 years ago
Read 2 more answers
Please help me answer question B!
skelet666 [1.2K]

Answer:

  • The program counter holds the memory address of the next instruction to be fetched from memory
  • The memory address register holds the address of memory from which data or instructions are to be fetched
  • The memory data register holds a copy of the memory contents transferred to or from the memory at the address in the memory address register
  • The accumulator holds the result of any logic or arithmetic operation

Step-by-step explanation:

The specific contents of any of these registers at any point in time <em>depends on the architecture of the computer</em>. If we make the assumption that the only interface registers connected to memory are the memory address register (MAR) and the memory data register (MDR), then <em>all memory transfers of any kind</em> will use both of these registers.

For execution of the instructions at addresses 01 through 03, the sequence of operations may go like this.

1. (Somehow) The program counter (PC) is set to 01.

2. The contents of the PC are copied to the MAR.

3. A Memory Read operation is performed, and the contents of memory at address 01 are copied to the MDR. (Contents are the LDA #11 instruction.)

4. The MDR contents are decoded (possibly after being transferred to an instruction register), and the value 11 is placed in the Accumulator.

5. The PC is incremented to 02.

6. The contents of the PC are copied to the MAR.

7. A Memory Read operation is performed, and the contents of memory at address 02 are copied to the MDR. (Contents are the SUB 05 instruction.)

8. The MDR contents are decoded and the value 05 is placed in the MAR.

9. A Memory Read operation is performed and the contents of memory at address 05 are copied to the MDR. (Contents are the value 3.)

10. The Accumulator contents are replaced by the difference of the previous contents (11) and the value in the MDR (3). The accumulator now holds the value 11 -3 = 8.

11. The PC is incremented to 03.

12. The contents of the PC are copied to the MAR.

13. A Memory Read operation is performed, and the contents of memory at address 03 are copied to the MDR. (Contents are the STO 06 instruction.)

14. The MDR contents are decoded and the value 06 is placed in the MAR.

15. The Accumulator value is placed in the MDR, and a Memory Write operation is performed. Memory address 06 now holds the value 8.

16. The PC is incremented to 04.

17. Instruction fetch and decoding continues. This program will go "off into the weeds", since there is no Halt instruction. Results are unpredictable.

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Note that decoding an instruction may result in several different data transfers and/or memory and/or arithmetic operations. All of this is usually completed before the next instruction is fetched.

In modern computers, memory contents may be fetched on the speculation that they will be used. Adjustments need to be made if the program makes a jump or if executing an instruction alters the data that was prefetched.

4 0
3 years ago
ASAP!!! Pls help me out :(
mr_godi [17]

The answer you are looking for is B. 70.69

3 0
3 years ago
Read 2 more answers
Mrs.Baca uses a phone card to call her relatives in Colombia.It costs her 45 cents to talk for 15 minutes.How long can she talk
zysi [14]
45/15= 3

So 1 minute=3 cents

75*3=225=2.25 converted

75 minutes=$2.25
6 0
3 years ago
Read 2 more answers
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